Although Java does not write pointers in plain text, pointers are still used.
dispatcherServlet.da(a); // Although the object a is written here, the reference

private A da(A a) { // 这里是一个指针对象, 指向上方传入的 那个对象
    a = new A("aaaaa"); // 这里只是改变了 指针对象 并没有修改 源对象.
    return a; // 如果你想获得新的对象, 那就取这个返回值来用吧.
}

Java is divided into objects, and reference objects (or pointers?) to this object.
Example:
Goods a in the warehouse, and the list of goods a qd_a, the list states which warehouse, which area and location in which goods a is located.
Someone changed the data on qd_a and pointed it to item b, but item a was not affected. What was affected was the list of qd_a.

This is what I understand. I don’t know if the explanation is clear. I hope you can correct me.

    private A da(A a) {
        a = new A("aaaaa");
        return a;
    }

Change a here to a member variable

Cprivate:
    A da(A &a) {
        a = new A("aaaaa");
        return a;
    }

Java does not pass by reference, which means that Java cannot let you replace the reference of the entire parameter. C/C++ has reference passing, but you must add the ampersand in front of the parameter.

Java passes by value, not by reference, but by passing the value of the pointer, so it looks like a reference. . .

Inside

a = new A("aaaaa");

This sentence cannot be reflected outside the function. If the value (address) of a is modified within the function, it is useless, but the content of a can be modified
For example, if there is

a.setA("aaaaa");

Then the value you print outside the function is "aaaaa"

Equivalent to C++ like this

void f(A *a) {
    a = new A("aaaaa");
}

void main() {
    a = new A("bb");
    f(a);
    cout << a.getA(); // "bb"
}

Java only passes by value, not by reference
Correct answer, when passing an object, you must remember that what you pass is not the memory address of the object itself, but the value of the object, and this value is the address referenced by the object

I feel like Java passes parameters on a case-by-case basis,
For container objects such as Map, the reference is passed instead of copying the value
The processing of the map container in the function will change to the original caller. This cannot be called passing by value...
It’s true that simple types are passed by value

Posts in 2009
http://ivan-pig.iteye.com/blog/422891
Can this solve lz’s question?

For non-primitive types, the reference is placed on the stack and the object itself is placed on the heap. The reference on the stack will be copied, but the object will not be copied.

If you want to implement reference transfer in Java, you can use boxing, that is, use an array to access.

private void foo(int[] iarr)
{
    int ival = iarr[0];
    // ...
    iarr[0] = ival;
}